Inverse of a selfpower matrix

Question

Suppose that $A$ is a matrix and we have $ A ^ 2 = A $ and also $ \gamma \neq 1 .$ Prove that $ I - \gamma A $ is invertible and find its inverse.

Answer 1

Hint:

Solve $(I-\gamma A)(I+\beta A)=I$ for $\beta$.

Another method when $|\gamma| $ is small enough:

$\dfrac 1{1-\gamma A}=1+\gamma A+\gamma^2 A^2+\gamma^3 A^3+\cdots$

$=1+A(\gamma+\gamma^2+\gamma^3+\cdots)=1+A\beta$.